To prove that $[p \land (A \lor B) \land (p \land A \implies q)] \implies q$ is not a tautology without a truth table $[p \land (A \lor B) \land (p \land A \implies q)] \implies q$
With the drawing of the truth table, it is apparent that the statement is false only when A is false and q is false.
I know for a fact that for the statement to be true,  


*

*$[p \land (A \lor B) \land (p \land A \implies q)]$ is true and $q$ is true

*$[p \land (A \lor B) \land (p \land A \implies q)]$ is false
I am unable to find a contraction is the case where  $[p \land (A \lor B) \land (p \land A \implies q)]$ is true and $q$ is true.

This is Q1-51(c) from Fundamentals of Mathematics: An Introduction to Proofs, Logic, Sets and Numbers btw.
Edit: Removed the incorrect portion
 A: As someone pointed out in the comments, in order to show that the statement is not a tautology, it is sufficient to find a proper assignment of $p,A,B$ and $q$ such that the statement (which we will call $r$) is false.
Now, lets inspect $r$. If you want $r$ to be false, then $s: [p \land (A \lor B) \land (p \land A \implies q)]$ must be true and $q$ must be false.
Now, given that $s$ must be true and has the form $ c \land d \land e$, then 


*

*$p$ must be true

*$A\lor B$ must be true

*$p \land A \implies q$ must be true


As we already know that $q$ must be false, the last one implies that $p \land A$ must be false. But, as $p$ must be true, then $A$ must be false.
Finally, our last assertion implies that $B$ must be true if we want $A\lor B$ to be true.
So there you have, in bold, your assignment that makes $r$ false. 
A: 
If $[p \land (A \lor B) \land (p \land A \to q)]$ is false, then $p$ is false, $A \lor B$ is false, $(p \land A \to q)$ is false.

No, then $p$ is false, OR $A \lor B$ is false, OR $(p \land A \to q)$ is false.  (deMorgan's Law).  
Also a sidetrack.

I am unable to find a contraction is the case where  $[p \land (A \lor B) \land (p \land A \implies q)]$ is true and $q$ is true.

To falsify the conditional statement what you need is an assignment where this antecedent is true, while the consequent ($q$) is false.
Ergo: If $\{p, (A \lor B), (p \land A \to q), \neg q\}$ is consistent then the statement is falsifiable.
Is it consistent?
A: One way to find a countermodel is to use a tree proof generator.  Any branch that does not close could be used to find a countermodel. 
Here is the output of one generator that also produces a countermodel if the branches of the tree do not all close:

One can visually check to see that this valuation would make the antecedent true, but the consequent false.

Tree Proof Generator. https://www.umsu.de/trees/
